Vector-Valued Functions and Space Curves

Definition of a Vector-Valued Function

Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. We can then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.

The parameter t can lie between two real numbers: \(a\le t\le b.\) Another possibility is that the value of t might take on all real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the value of t. We often use t as a parameter because t can represent time.

illustrates an important concept. The domain of a vector-valued function consists of real numbers. The domain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors. Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.

Condensed: the full section is in OpenStax Calculus Volume 3.

Graphing Vector-Valued Functions

Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point), if we move in a specific direction for a specific distance, we arrive at a second point. This represents the terminal point of the vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates of the terminal point.

A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valued function, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees the uniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. The graph of a vector-valued function of the form \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}\) consists of the set of all \((t,\ \text{r}(t)),\) and the path it traces is called a plane curve. The graph of a vector-valued function of the form \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}\) consists of the set of all \((t,\ \text{r}(t)),\) and the path it traces is called a space curve. Any representation of a plane curve or space curve using a vector-valued function is called a vector parameterization of the curve.

You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curve b is a so-called reparameterization of the function describing curve a. In fact, any curve has an infinite number of reparameterizations; for example, we can replace t with \(2t\) in any of the three previous curves without changing the shape of the curve. The interval over which t is defined may change, but that is all. We return to this idea later in this chapter when we study arc-length parameterization.

As mentioned, the name of the shape of the curve of the graph in c. is a helix (). The curve resembles a spring, with a circular cross-section looking down along the z-axis. It is possible for a helix to be elliptical in cross-section as well. For example, the vector-valued function \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+3\ \text{sin}\ t\ \text{j}+t\ \text{k}\) describes an elliptical helix. The projection of this helix into the \(x,y\text{-plane}\) is an ellipse. Last, the arrows in the graph of this helix indicate the orientation of the curve as t progresses from 0 to \(4\pi .\)

Condensed: the full section is in OpenStax Calculus Volume 3.

Limits and Continuity of a Vector-Valued Function

We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.

This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:

In the following example, we show how to calculate the limit of a vector-valued function.

Example

Try it.

For each of the following vector-valued functions, calculate \(\underset{t\to 3}{\text{lim}}\ \text{r}(t)\) for

1. \(\text{r}(t)=({t}^{2}-3t+4)\ \text{i}+(4t+3)\ \text{j}\)
2. \(\text{r}(t)=\frac{2t-4}{t+1}\ \text{i}+\frac{t}{{t}^{2}+1}\ \text{j}+(4t-3)\ \text{k}\)

Solution

1. Use and substitute the value \(t=3\) into the two component expressions:
   \[\begin{array}{ll}\underset{t\to 3}{\text{lim}}\ \text{r}(t) & =\underset{t\to 3}{\text{lim}}[({t}^{2}-3t+4)\ \text{i}+(4t+3)\ \text{j}] \\ & =[\underset{t\to 3}{\text{lim}}({t}^{2}-3t+4)]\ \text{i}+[\underset{t\to 3}{\text{lim}}(4t+3)]\ \text{j} \\ & =4\ \text{i}+15\ \text{j}.\end{array}\]
2. Use and substitute the value \(t=3\) into the three component expressions:
   \[\begin{array}{ll}\underset{t\to 3}{\text{lim}}\ \text{r}(t) & =\underset{t\to 3}{\text{lim}}(\frac{2t-4}{t+1}\ \text{i}+\frac{t}{{t}^{2}+1}\ \text{j}+(4t-3)\ \text{k}) \\ & =[\underset{t\to 3}{\text{lim}}(\frac{2t-4}{t+1})]\ \text{i}+[\underset{t\to 3}{\text{lim}}(\frac{t}{{t}^{2}+1})]\ \text{j}+[\underset{t\to 3}{\text{lim}}(4t-3)]\ \text{k} \\ & =\frac{1}{2}\ \text{i}+\frac{3}{10}\ \text{j}+9\ \text{k}.\end{array}\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Key Concepts

• A vector-valued function is a function of the form \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}\) or \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k},\) where the component functions f, g, and h are real-valued functions of the parameter t.
• The graph of a vector-valued function of the form \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}\) is called a plane curve. The graph of a vector-valued function of the form \(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k}\) is called a space curve.
• It is possible to represent an arbitrary plane curve by a vector-valued function.
• To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.

Key Equations

Vector-valued function	\(\text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}\ \text{or}\ \text{r}(t)=f(t)\ \text{i}+g(t)\ \text{j}+h(t)\ \text{k},\ \text{or}\ \text{r}(t)=〈f(t),g(t)〉\ \text{or}\ \text{r}(t)=〈f(t),g(t),h(t)〉\)
Limit of a vector-valued function	\(\underset{t\to a}{\text{lim}}\ \text{r}(t)=[\underset{t\to a}{\text{lim}}f(t)]\ \text{i}+[\underset{t\to a}{\text{lim}}g(t)]\ \text{j}\ \text{or}\ \underset{t\to a}{\text{lim}}\ \text{r}(t)=[\underset{t\to a}{\text{lim}}f(t)]\ \text{i}+[\underset{t\to a}{\text{lim}}g(t)]\ \text{j}+[\underset{t\to a}{\text{lim}}h(t)]\ \text{k}\)

Vector-Valued Functions and Space Curves

Find the limit of the following vector-valued functions at the indicated value of t.

Find the domain of the vector-valued functions.

Let \(\text{r}(t)=〈\text{cos}\ t,t,\text{sin}\ t〉\) and use it to answer the following questions.

Eliminate the parameter t, write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions.

Use a graphing utility to sketch each of the following vector-valued functions:

Consider the curve described by the vector-valued function \(\text{r}(t)=(50{e}^{\text{-}t}\text{cos}\ t)\ \text{i}+(50{e}^{\text{-}t}\text{sin}\ t)\ \text{j}+(5-5{e}^{\text{-}t})\ \text{k}.\)

今度はあなたの番 計算機では解けませんが、計算可能です。下の一つを試してみましょう。もしくは自分で入力してください。

⌨ 入力 ✎ 描いて

数学の問題

π 解く

練習 (40)

まず紙で解いてみてください。答えを表示して確認できます。検証済みの問題は、ソルバーで全ステップを開けます。

1. For each of the following vector-valued functions, evaluate \(\text{r}(0),\ \text{r}(\frac{\pi }{2}),\text{and}\ \text{r}(\frac{2\pi }{3}).\) Do any of these functions have domain restrictions?

   1. \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+3\ \text{sin}\ t\ \text{j}\)
   2. \(\text{r}(t)=3\ \text{tan}\ t\ \text{i}+4\ \text{sec}\ t\ \text{j}+5t\ \text{k}\)
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   答えを表示

   1. To calculate each of the function values, substitute the appropriate value of t into the function:
      \[\begin{array}{lll}\text{r}(0) & = & 4\ \text{cos}(0)\ \text{i}+3\ \text{sin}(0)\ \text{j} \\ & = & 4\ \text{i}+0\ \text{j}=4\ \text{i} \\ \text{r}(\frac{\pi }{2}) & = & 4\ \text{cos}(\frac{\pi }{2})\ \text{i}+3\ \text{sin}(\frac{\pi }{2})\ \text{j} \\ & = & 0\ \text{i}+3\ \text{j}=3\ \text{j} \\ \text{r}(\frac{2\pi }{3}) & = & 4\ \text{cos}(\frac{2\pi }{3})\ \text{i}+3\ \text{sin}(\frac{2\pi }{3})\ \text{j} \\ & = & 4(-\frac{1}{2})\ \text{i}+3(\frac{\sqrt{3}}{2})\ \text{j}=-2\ \text{i}+\frac{3\sqrt{3}}{2}\ \text{j}.\end{array}\]
      To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is \(f(t)=4\ \text{cos}\ t\) and the second component function is \(g(t)=3\ \text{sin}\ t.\) Neither of these functions has a domain restriction, so the domain of \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+3\ \text{sin}\ t\ \text{j}\) is all real numbers.
   2. To calculate each of the function values, substitute the appropriate value of t into the function:
      \[\begin{array}{lll}\ \text{r}(0) & = & 3\ \text{tan}(0)\ \text{i}+4\ \text{sec}(0)\ \text{j}+5(0)\ \text{k} \\ & = & 0\ \text{i}+4\ \text{j}+0\ \text{k}=4\ \text{j} \\ \text{r}(\frac{\pi }{2}) & = & 3\ \text{tan}(\frac{\pi }{2})\ \text{i}+4\ \text{sec}(\frac{\pi }{2})\ \text{j}+5(\frac{\pi }{2})\ \text{k},\text{which does not exist} \\ \text{r}(\frac{2\pi }{3}) & = & 3\ \text{tan}(\frac{2\pi }{3})\ \text{i}+4\ \text{sec}(\frac{2\pi }{3})\ \text{j}+5(\frac{2\pi }{3})\ \text{k} \\ & = & 3(-\sqrt{3})\ \text{i}+4(-2)\ \text{j}+\frac{10\pi }{3}\ \text{k} \\ & = & -3\sqrt{3}\ \text{i}-8\ \text{j}+\frac{10\pi }{3}\ \text{k}.\end{array}\]
      To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is \(f(t)=3\ \text{tan}\ t,\) the second component function is \(g(t)=4\ \text{sec}\ t,\) and the third component function is \(h(t)=5t.\) The first two functions are not defined for odd multiples of \(\pi \text{/}2,\) so the function is not defined for odd multiples of \(\pi \text{/}2.\) Therefore, \(\text{dom}(\ \text{r}(t))=\{t\ |t\ne \frac{(2n+1)\pi }{2}\},\) where n is any integer.

2. For the vector-valued function \(\text{r}(t)=({t}^{2}-3t)\ \text{i}+(4t+1)\ \text{j},\) evaluate \(\text{r}(0),\ \text{r}(1),\text{and}\ \text{r}(-4).\) Does this function have any domain restrictions?

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   答えを表示

   \(\text{r}(0)=\text{j},\ \text{r}(1)=-2\ \text{i}+5\ \text{j},\ \text{r}(-4)=28\ \text{i}-15\ \text{j}\)

   The domain of \(\text{r}(t)=({t}^{2}-3t)\ \text{i}+(4t+1)\ \text{j}\) is all real numbers.

3. Create a graph of each of the following vector-valued functions:

   1. The plane curve represented by \(\text{r}(t)=4\ \text{cos}\ t\ \text{i}+3\ \text{sin}\ t\ \text{j},\) \(0\le t\le 2\pi\)
   2. The plane curve represented by \(\text{r}(t)=4\ \text{cos}\ 2t\text{i}+3\ \text{sin}\ 2\ t\text{j},\) \(0\le t\le \pi\)
   3. The space curve represented by \(\text{r}(t)=4\text{cos}\ t\ \text{i}+4\text{sin}\ t\ \text{j}+t\ \text{k},\) \(0\le t\le 4\pi\)
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   答えを表示

   1. As with any graph, we start with a table of values. We then graph each of the vectors in the second column of the table in standard position and connect the terminal points of each vector to form a curve (). This curve turns out to be an ellipse centered at the origin.
      

      t	\(\text{r}(t)\)	t	\(\text{r}(t)\)
      0	\(4\ \text{i}\)	\(\pi\)	\(-4\ \text{i}\)
      \(\frac{\pi }{4}\)	\(2\sqrt{2}\ \text{i}+\frac{3\sqrt{2}}{2}\ \text{j}\)	\(\frac{5\pi }{4}\)	\(-2\sqrt{2}\ \text{i}-\frac{3\sqrt{2}}{2}\ \text{j}\)
      \(\frac{\pi }{2}\)	\(3\ \text{j}\)	\(\frac{3\pi }{2}\)	\(-3\ \text{j}\)
      \(\frac{3\pi }{4}\)	\(-2\sqrt{2}\ \text{i}+\frac{3\sqrt{2}}{2}\ \text{j}\)	\(\frac{7\pi }{4}\)	\(2\sqrt{2}\ \text{i}-\frac{3\sqrt{2}}{2}\ \text{j}\)
      \(2\pi\)	\(4\ \text{i}\)

   2. The table of values for \(\text{r}(t)=4\text{cos}\ 2t\text{i}+3\ \text{sin}\ 2\ t\text{j},\) \(0\le t\le \pi\) is as follows:
      

      t	\(\text{r}(t)\)	t	\(\text{r}(t)\)
      0	\(4\ \text{i}\)	\(\frac{\pi }{2}\)	\(-4\ \text{i}\)
      \(\frac{\pi }{8}\)	\(2\sqrt{2}\ \text{i}+\frac{3\sqrt{2}}{2}\ \text{j}\)	\(\frac{5\pi }{8}\)	\(-2\sqrt{2}\ \text{i}-\frac{3\sqrt{2}}{2}\ \text{j}\)
      \(\frac{\pi }{4}\)	\(3\ \text{j}\)	\(\frac{3\pi }{4}\)	\(-3\ \text{j}\)
      \(\frac{3\pi }{8}\)	\(-2\sqrt{2}\ \text{i}+\frac{3\sqrt{2}}{2}\ \text{j}\)	\(\frac{7\pi }{8}\)	\(2\sqrt{2}\ \text{i}-\frac{3\sqrt{2}}{2}\ \text{j}\)
      		\(\pi\)	\(4\ \text{i}\)

      
      The graph of this curve is also an ellipse centered at the origin.
   3. We go through the same procedure for a three-dimensional vector function.
      

      t	\(\text{r}(t)\)	t	\(\text{r}(t)\)
      0	\(4\ \text{i}\)	\(\pi\)	\(-4\ \text{j}+\pi \ \text{k}\)
      \(\frac{\pi }{4}\)	\(2\sqrt{2}\ \text{i}+2\sqrt{2}\ \text{j}+\frac{\pi }{4}\ \text{k}\)	\(\frac{5\pi }{4}\)	\(-2\sqrt{2}\ \text{i}-2\sqrt{2}\ \text{j}+\frac{5\pi }{4}\ \text{k}\)
      \(\frac{\pi }{2}\)	\(4\ \text{j}+\frac{\pi }{2}\ \text{k}\)	\(\frac{3\pi }{2}\)	\(-4\ \text{j}+\frac{3\pi }{2}\ \text{k}\)
      \(\frac{3\pi }{4}\)	\(-2\sqrt{2}\ \text{i}+2\sqrt{2}\ \text{j}+\frac{3\pi }{4}\ \text{k}\)	\(\frac{7\pi }{4}\)	\(2\sqrt{2}\ \text{i}-2\sqrt{2}\ \text{j}+\frac{7\pi }{4}\ \text{k}\)
      \(2\pi\)	\(4\ \text{i}+2\pi \ \text{k}\)

      
      The values then repeat themselves, except for the fact that the coefficient of k is always increasing (). This curve is called a helix. Notice that if the k component is eliminated, then the function becomes \(\text{r}(t)=\text{cos}\ t\ \text{i}+\text{sin}\ t\ \text{j},\) which is a unit circle centered at the origin.

4. Create a graph of the vector-valued function \(\text{r}(t)=({t}^{2}-1)\ \text{i}+(2t-3)\ \text{j},\) \(0\le t\le 3.\)

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5. For each of the following vector-valued functions, calculate \(\underset{t\to 3}{\text{lim}}\ \text{r}(t)\) for

   1. \(\text{r}(t)=({t}^{2}-3t+4)\ \text{i}+(4t+3)\ \text{j}\)
   2. \(\text{r}(t)=\frac{2t-4}{t+1}\ \text{i}+\frac{t}{{t}^{2}+1}\ \text{j}+(4t-3)\ \text{k}\)
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   答えを表示

   1. Use and substitute the value \(t=3\) into the two component expressions:
      \[\begin{array}{ll}\underset{t\to 3}{\text{lim}}\ \text{r}(t) & =\underset{t\to 3}{\text{lim}}[({t}^{2}-3t+4)\ \text{i}+(4t+3)\ \text{j}] \\ & =[\underset{t\to 3}{\text{lim}}({t}^{2}-3t+4)]\ \text{i}+[\underset{t\to 3}{\text{lim}}(4t+3)]\ \text{j} \\ & =4\ \text{i}+15\ \text{j}.\end{array}\]
   2. Use and substitute the value \(t=3\) into the three component expressions:
      \[\begin{array}{ll}\underset{t\to 3}{\text{lim}}\ \text{r}(t) & =\underset{t\to 3}{\text{lim}}(\frac{2t-4}{t+1}\ \text{i}+\frac{t}{{t}^{2}+1}\ \text{j}+(4t-3)\ \text{k}) \\ & =[\underset{t\to 3}{\text{lim}}(\frac{2t-4}{t+1})]\ \text{i}+[\underset{t\to 3}{\text{lim}}(\frac{t}{{t}^{2}+1})]\ \text{j}+[\underset{t\to 3}{\text{lim}}(4t-3)]\ \text{k} \\ & =\frac{1}{2}\ \text{i}+\frac{3}{10}\ \text{j}+9\ \text{k}.\end{array}\]

6. Calculate \(\underset{t\to -2}{\text{lim}}\ \text{r}(t)\) for the function \(\text{r}(t)=\sqrt{{t}^{2}-3t-1}\ \text{i}+(4t+3)\ \text{j}+\text{sin}\ \frac{(t+1)\pi }{2}\ \text{k}.\)

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   答えを表示

   \(\underset{t\to -2}{\text{lim}}\ \text{r}(t)=3\ \text{i}-5\ \text{j}-\text{k}\)

7. Give the component functions \(x=f(t)\) and \(y=g(t)\) for the vector-valued function \(\text{r}(t)=3\ \text{sec}\ t\ \text{i}+2\ \text{tan}\ t\ \text{j}.\)

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   答えを表示

   \(f(t)=3\ \text{sec}\ t,g(t)=2\ \text{tan}\ t\)

8. Given \(\text{r}(t)=3\ \text{sec}\ t\ \text{i}+2\ \text{tan}\ t\ \text{j},\) find the following values (if possible).

   1. \(\text{r}(\frac{\pi }{4})\)
   2. \(\text{r}(\pi )\)
   3. \(\text{r}(\frac{\pi }{2})\)
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9. Sketch the curve of the vector-valued function \(\text{r}(t)=3\ \text{sec}\ t\ \text{i}+2\ \text{tan}\ t\ \text{j}\) and give the orientation of the curve. Sketch asymptotes as a guide to the graph.

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10. Evaluate \(\underset{t\to 0}{\text{lim}}〈{e}^{t}\ ,\frac{\text{sin}\ t}{t}\ ,{e}^{\text{-}t}〉.\)

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11. Given the vector-valued function \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t〉,\) find the following values:

    1. \(\underset{t\to \frac{\pi }{3}}{\text{lim}}\ \text{r}(t)\)
    2. \(\text{r}(\frac{\pi }{3})\)
    3. Is \(\text{r}(t)\) continuous at \(t=\frac{\pi }{3}?\)
    4. Graph \(\text{r}(t).\)
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    答えを表示

    a. \(〈\frac{1}{2},\frac{\sqrt{3}}{2}〉,\) b. \(〈\frac{1}{2},\frac{\sqrt{3}}{2}〉,\) c. Yes, the limit as t approaches \(\pi \text{/}3\) is equal to \(\text{r}(\pi \text{/}3),\) d.

12. Given the vector-valued function \(\text{r}(t)=〈t,{t}^{2}+1〉,\) find the following values:

    1. \(\underset{t\to -3}{\text{lim}}\ \text{r}(t)\)
    2. \(\text{r}(-3)\)
    3. Is \(\text{r}(t)\) continuous at \(t=-3?\)
    4. \(\text{r}(t+2)-\text{r}(t)\)
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13. Let \(\text{r}(t)={e}^{t}\ \text{i}+\text{sin}\ t\ \text{j}+\text{ln}\ t\ \text{k}.\) Find the following values:

    1. \(\text{r}(\frac{\pi }{4})\)
    2. \(\underset{t\to \pi \text{/}4}{\text{lim}}\ \text{r}(t)\)
    3. Is \(\text{r}(t)\) continuous at \(t=\frac{\pi }{4}?\)
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    答えを表示

    a. \(〈{e}^{\pi \text{/}4},\frac{\sqrt{2}}{2},\text{ln}(\frac{\pi }{4})〉;\) b. \(〈{e}^{\pi \text{/}4},\frac{\sqrt{2}}{2},\text{ln}(\frac{\pi }{4})〉;\) c. Yes

14. \(\underset{t\to 4}{\text{lim}}〈\sqrt{t-3},\frac{\sqrt{t}-2}{t-4},\text{tan}(\frac{\pi }{t})〉\)

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15. \(\underset{t\to \pi \text{/}2}{\text{lim}}\ \text{r}(t)\) for \(\text{r}(t)={e}^{t}\ \text{i}+\text{sin}\ t\ \text{j}+\text{ln}\ t\ \text{k}\)

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    答えを表示

    \(〈{e}^{\pi \text{/}2},1,\text{ln}(\frac{\pi }{2})〉\)

16. \(\underset{t\to \infty }{\text{lim}}〈{e}^{-2t},\frac{2t+3}{3t-1},\text{arctan}(2t)〉\)

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17. \(\underset{t\to {e}^{2}}{\text{lim}}〈t\ \text{ln}(t),\frac{\text{ln}\ t}{{t}^{2}},\sqrt{\text{ln}\ ({t}^{2})}〉\)

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    答えを表示

    \(2{e}^{2}\ \text{i}+\frac{2}{{e}^{4}}\ \text{j}+2\ \text{k}\)

18. \(\underset{t\to \pi \text{/}6}{\text{lim}}〈{\text{cos}}^{2}t,{\text{sin}}^{2}t,1〉\)

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19. \(\underset{t\to \infty }{\text{lim}}\ \text{r}(t)\) for \(\text{r}(t)=2{e}^{\text{-}t}\ \text{i}+{e}^{\text{-}t}\ \text{j}+\text{ln}(t-1)\ \text{k}\)

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    答えを表示

    The limit does not exist because the limit of \(\text{ln}(t-1)\) as t approaches infinity does not exist.

20. Describe the curve defined by the vector-valued function \(\text{r}(t)=(1+t)\ \text{i}+(2+5t)\ \text{j}+(-1+6t)\ \text{k}.\)

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21. Domain: \(\text{r}(t)=〈{t}^{2},\text{tan}\ t,\text{ln}\ t〉\)

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    答えを表示

    \(t>0,t\ne (2k+1)\frac{\pi }{2},\) where k is an integer

22. Domain: \(\text{r}(t)=〈{t}^{2},\sqrt{t-3},\frac{3}{2t+1}〉\)

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23. Domain: \(\text{r}(t)=〈\text{csc}(t),\frac{1}{\sqrt{t-3}},\text{ln}(t-2)〉\)

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    答えを表示

    \(t>3,t\ne n\pi ,\) where n is an integer

24. For what values of t is \(\text{r}(t)\) continuous?

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25. Sketch the graph of \(\text{r}(t).\)

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26. Find the domain of \(\text{r}(t)=2{e}^{\text{-}t}\ \text{i}+{e}^{\text{-}t}\ \text{j}+\text{ln}(t-1)\ \text{k}.\)

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27. For what values of t is \(\text{r}(t)=2{e}^{\text{-}t}\ \text{i}+{e}^{\text{-}t}\ \text{j}+\text{ln}(t-1)\ \text{k}\) continuous?

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    答えを表示

    All t such that \(t\in (1,\infty )\)

28. \(\text{r}(t)=2t\ \text{i}+{t}^{2}\ \text{j}\) (Hint: Let \(x=2t\) and \(y={t}^{2}.\) Solve the first equation for x in terms of t and substitute this result into the second equation.)

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29. \(\text{r}(t)={t}^{3}\ \text{i}+2t\ \text{j}\)

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    答えを表示

    \(y=2\sqrt[3]{x},\) a variation of the cube-root function

30. \(\text{r}(t)=2(\text{sinh}\ t)\ \text{i}+2(\text{cosh}\ t)\ \text{j},t>0\)

    ステップごとに解く ヒント

31. \(\text{r}(t)=3(\text{cos}\ t)\ \text{i}+3(\text{sin}\ t)\ \text{j}\)

    ステップごとに解く ヒント

    答えを表示

    \({x}^{2}+{y}^{2}=9,\) a circle centered at \((0,0)\) with radius 3, and a counterclockwise orientation

32. \(\text{r}(t)=〈3\ \text{sin}\ t,3\ \text{cos}\ t〉\)

    ステップごとに解く ヒント

33. [T] \(\text{r}(t)=\left(2\ {\text{cos}}^{2}\ t\right)\ \text{i}+(2-\sqrt{t})\ \text{j}\)

    ステップごとに解く ヒント

34. [T] \(\text{r}(t)=〈{e}^{\text{cos}(3t)},{e}^{\text{-}\text{sin}(t)}〉\)

    ステップごとに解く ヒント

35. [T] \(\text{r}(t)=〈2-\text{sin}(2t),3+2\ \text{cos}\ t〉\)

    ステップごとに解く ヒント

    答えを表示

    Find a vector-valued function that traces out the given curve in the indicated direction.

36. \(4{x}^{2}+9{y}^{2}=36;\) clockwise and counterclockwise

    ステップごとに解く ヒント

37. \(\text{r}(t)=〈t,{t}^{2}〉;\) from left to right

    ステップごとに解く ヒント

    答えを表示

    For left to right, \(y={x}^{2},\) where t increases

38. The line through P and Q where P is \((1,4,-2)\) and Q is \((3,9,6)\)

    ステップごとに解く ヒント

39. What is the initial point of the path corresponding to \(\text{r}(0)?\)

    ステップごとに解く ヒント

    答えを表示

    \((50,0,0)\)

40. What is \(\underset{t\to \infty }{\text{lim}}\ \text{r}(t)?\)

    ステップごとに解く ヒント

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ここで使用された記号

どの記号をタップしても、定義、画像、それぞれの文字の意味が表示されます。

\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.\inftyinfinity
Not a number: "grows without bound" in limits and intervals.\pipi
Ratio of a circle's circumference to its diameter, 3.14159…iimaginary unit
i² = −1.\leqless/greater than or equal
Inequalities that allow equality; < and > exclude it.\supsupremum, infimum
Least upper bound, greatest lower bound.\frac{\partial f}{\partial x}partial derivative
Derivative with respect to x, holding the other variables fixed.\nabla fgradient (nabla, del)
Vector of partial derivatives; points uphill.\int f(x)integral
Antiderivative (indefinite) or signed area from a to b (definite).\iintdouble / contour integral
Integral over a region of the plane; integral around a closed curve.\mathbf{v}vector
A quantity with magnitude and direction; a column of numbers.\mathbf{u} \cdot \mathbf{v}dot product, norm
Σ u_i v_i; the length of v, √(v·v).

どうやって: Vector-Valued Functions and Space Curves

1. Write the general equation of a vector-valued function in component form and unit-vector form.
2. Recognize parametric equations for a space curve.
3. Describe the shape of a helix and write its equation.
4. Define the limit of a vector-valued function.
5. To calculate each of the function values, substitute the appropriate value of
6. To calculate each of the function values, substitute the appropriate value of
7. The plane curve represented by
8. The plane curve represented by

よくある質問

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen: the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). ここでは要約して説明し直しています。誤りは私たちの責任です。 ソースは非商用利用のためにライセンスされているため、このレッスンは無料で、売ることはなく、同じライセンスの下で共有されている。

More in Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems